Question

On the interval $[0,1]$, the function $x^{25}(1-x{)}^{75}$ takes its maximum value at the point

• A. $0$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

Answer: (B)

Let $f\left(x\right)=x^{25}{\left(1-x\right)}^{75},x\in \left[0,1\right]$
$\Rightarrow f^{\prime }\left(x\right)=25x^{24}{\left(1-x\right)}^{75}-75x^{25}{\left(1-x\right)}^{74}$
$=25x^{24}{\left(1-x\right)}^{74}\left(1-x\right)-3x$
$=25x^{24}{\left(1-x\right)}^{74}\left(1-4x\right)$

We can see that $f^{\prime }\left(x\right)$ is positive for $x<\frac{1}{4}$
and $f^{\prime }\left(x\right)$ is negative for $x>\frac{1}{4}$.
Hence, $f\left(x\right)$ attains maximum at $x=\frac{1}{4}.$